10.6 Area

A diagram of a baseball field with its measurements marked. — **Figure 10.103** The area of a regulation baseball diamond must adhere to specific measurements to be legal.The area of a regulation baseball diamond must adhere to specific measurements to be legal. (credit: "Diagram of Regulation Diamond" by Erica Fischer from "Baseball" The World Book, 1920/Flickr, Public Domain, CC BY 2.0)

Learning Objectives

After completing this section, you should be able to:

Calculate the area of triangles.
Calculate the area of quadrilaterals.
Calculate the area of other polygons.
Calculate the area of circles.

Some areas carry more importance than other areas. Did you know that in a baseball game, when the player hits the ball and runs to first base that he must run within a 6-foot wide path? If he veers off slightly to the right, he is out. In other words, a few inches can be the difference in winning or losing a game. Another example is real estate. On Manhattan Island, one square foot of real estate is worth far more than real estate in practically any other area of the country. In other words, we place a value on area. As the context changes, so does the value.

Area refers to a region measured in square units, like a square mile or a square foot. For example, to purchase tile for a kitchen floor, you would need to know how many square feet are needed because tile is sold by the square foot. Carpeting is sold by the square yard. As opposed to linear measurements like perimeter, which in in linear units. For example, fencing is sold in linear units, a linear foot or yard. Linear dimensions refer to an outline or a boundary. Square units refer to the area within that boundary. Different items may have different units, but either way, you must know the linear dimensions to calculate the area.

Many geometric shapes have formulas for calculating areas, such as triangles, regular polygons, and circles. To calculate areas for many irregular curves or shapes, we need calculus. However, in this section, we will only look at geometric shapes that have known area formulas. The notation for area, as mentioned, is in square units and we write sq in or sq cm, or use an exponent, such as ${in}^{2}$ or ${cm}^{2} .$ Note that linear measurements have no exponent above the units or we can say that the exponent is 1.

Area of Triangles

The formula for the area of a triangle is given as follows.

For example, consider the triangle in Figure 10.104.

A right triangle with its legs marked 5 centimeters and 4 centimeters. — **Figure 10.104** Triangle 1

The base measures 4 cm and the height measures 5 cm. Using the formula, we can calculate the area:

$A = \frac{1}{2} (4) (5) = \frac{1}{2} (20) = 10 {cm}^{2}$

In Figure 10.105, the triangle has a base equal to 7 cm and a height equal to 3.5 cm. Notice that we can only find the height by dropping a perpendicular to the base. The area is then

$A = \frac{1}{2} (7) (3.5) = 12.25 {cm}^{2} .$

A triangle with its base marked b equals 7 centimeters and height marked h equals 3.5 centimeters. — **Figure 10.105** Triangle 2

Area of Quadrilaterals

A quadrilateral is a four-sided polygon with four vertices. Some quadrilaterals have either one or two sets of parallel sides. The set of quadrilaterals include the square, the rectangle, the parallelogram, the trapezoid, and the rhombus. The most common quadrilaterals are the square and the rectangle.

Square

In Figure 10.107, a $12 in times 12 in$ grid is represented with twelve $1 in times 1 in$ squares across each row, and twelve $1 in times 1 in$ squares down each column. If you count the little squares, the sum equals 144 squares. Of course, you do not have to count little squares to find area—we have a formula. Thus, the formula for the area of a square, where $s = length of a side$ , is $A = s \cdot s$ . The area of the square in Figure 10.107 is $A = 12 in \times 12 in = 144 {in}^{2} .$

Rectangle

Similarly, the area for a rectangle is found by multiplying length times width. The rectangle in Figure 10.108 has width equal to 5 in and length equal to 12 in. The area is $A = 5 (12) = 60 {in}^{2} .$

Many everyday applications require the use of the perimeter and area formulas. Suppose you are remodeling your home and you want to replace all the flooring. You need to know how to calculate the area of the floor to purchase the correct amount of tile, or hardwood, or carpet. If you want to paint the rooms, you need to calculate the area of the walls and the ceiling to know how much paint to buy, and the list goes on. Can you think of other situations where you might need to calculate area?

Parallelogram

The area of a parallelogram can be found using the formula for the area of a triangle. Notice in Figure 10.109, if we cut a diagonal across the parallelogram from one vertex to the opposite vertex, we have two triangles. If we multiply the area of a triangle by 2, we have the area of a parallelogram:

$\begin{matrix} A & = & 2 (\frac{1}{2} b h) \\ A & = & b h \end{matrix}$

A parallelogram with its base marked b and height marked h. — **Figure 10.109** Area of a Parallelogram

For example, if we have a parallelogram with the base be equal to 10 inches and the height equal to 5 inches, the area will be $A = (10) (5) = 50 {in}^{2} .$

Example 5 Finding the Area of a Parallelogram Park

The boundaries of a city park form a parallelogram (Figure 10.111). The park takes up one city block, which is contained by two sets of parallel streets. Each street measures 55 yd long. The perpendicular distance between streets is 39 yd. How much sod, sold by the square foot, should the city purchase to cover the entire park and how much will it cost? The sod is sold for $0.50 per square foot, installation is $1.50 per square foot, and the cost of the equipment for the day is $100.

Figure 10.111

Step 1: As sod is sold by the square foot, the first thing we have to do is translate the measurements of the park from yards to feet. There are 3 ft to a yard, so 55 yd is equal to 165 ft, and 39 yd is equal to 117 ft.

Step 2: The park has the shape of a parallelogram, and the formula for the area is $A = b h$ :

$A = 165 (117) = 19,305 {ft}^{2} .$

Step 3: The city needs to purchase $19, 305 ft squared$ of sod. The cost will be $0.50 per square foot for the sod and $1.50 per square foot for installation, plus $100 for equipment:

$\begin{matrix} 19,305 ($ 0.50) + 19,305 ($ 1.50) & = & $ 9,652.5 + $ 28,957.5 + $ 100.00 \\ = & $ 38,710.00 \end{matrix}$

Trapezoid

Another quadrilateral is the trapezoid. A trapezoid has one set of parallel sides or bases. The formula for the area of a trapezoid with parallel bases $a$ and $b$ and height $h$ is given here.

For example, find the area of the trapezoid in Figure 10.112 that has base $a$ equal to 8 cm, base $b$ equal to 6 cm, and height equal to 6 cm.

A trapezoid with the bottom and top bases marked a and b. The height is marked h. — **Figure 10.112** Area of a Trapezoid

The area is $A = \frac{1}{2} (6) (6 + 8) = 42 {cm}^{2}$ .

Rhombus

The rhombus has two sets of parallel sides. To find the area of a rhombus, there are two formulas we can use. One involves determining the measurement of the diagonals.

For our purposes here, we will use the formula that uses diagonals. For example, if the area of a rhombus is $220 cm squared,$ and the measure of $d sub 2 equals 11,$ find the measure of $d sub 1.$ To solve this problem, we input the known values into the formula and solve for the unknown. See Figure 10.114.

A rhombus with two diagonal lines labeled d subscript 1 and d subscript 2. The line, d subscript 2 measures 11. — **Figure 10.114** Area of a Rhombus

We have that

$\begin{matrix} 220 & = & \frac{11 d_{1}}{2} \\ 220 (2) & = & 11 d_{1} \\ \frac{440}{11} & = & d_{1} = 40 \end{matrix}$

Area of Polygons

To find the area of a regular polygon, we need to learn about a few more elements. First, the apothem $a$ of a regular polygon is a line segment that starts at the center and is perpendicular to a side. The radius $r$ of a regular polygon is also a line segment that starts at the center but extends to a vertex. See Figure 10.116.

A hexagon with its radius marked r and apothem marked a. — **Figure 10.116** Apothem and Radius of a Polygon

For example, consider the regular hexagon shown in Figure 10.117 with a side length of 4 cm, and the apothem measures $a = 2 \sqrt{3} .$

A hexagon with its center marked C. Each side measures 4 centimeters. The apothem is marked a equals 2 times square root of 3. — **Figure 10.117** Area of a Hexagon

We have the perimeter, $p = 6 (4) = 24 cm .$ We have the apothem as $a = 2 \sqrt{3} .$ Then, the area is:

$\begin{matrix} A & = & \frac{1}{2} (2 \sqrt{3}) (24) \\ = & 24 \sqrt{3} = 41.57 {cm}^{2} \end{matrix}$

Changing Units

Often, we have the need to change the units of one or more items in a problem to find a solution. For example, suppose you are purchasing new carpet for a room measured in feet, but carpeting is sold in terms of yards. You will have to convert feet to yards to purchase the correct amount of carpeting. Or, you may need to convert centimeters to inches, or feet to meters. In each case, it is essential to use the correct equivalency.

Area of Circles

Just as the circumference of a circle includes the number $π,$ so does the formula for the area of a circle. Recall that $π$ is a non-terminating, non-repeating decimal number: $π equals 3.14159 and so on$ . It represents the ratio of the circumference to the diameter, so it is a critical number in the calculation of circumference and area.

For example, to find the of the circle with radius equal to 3 cm, as shown in Figure 10.119, is found using the formula $A = π r^{2} .$

A circle with its radius, r marked 3 centimeters. — **Figure 10.119** Circle with Radius 3

We have

$\begin{matrix} A & = & π r^{2} \\ = & π {(3)}^{2} \\ = & 9 π = 28.27 {cm}^{2} \end{matrix}$

Example 13 Applying Area to the Real World

You want to purchase a tinted film, sold by the square foot, for the window in Figure 10.120. (This problem should look familiar as we saw it earlier when calculating circumference.) The bottom part of the window is a rectangle, and the top part is a semicircle. Find the area and calculate the amount of film to purchase.

Figure 10.120

First, the rectangular portion has $A = 5 (10) = 50 {ft}^{2}$ . For the top part, we have a semicircle with a diameter of 5 ft, so the radius is 2.5 ft. We want one half of the area of a circle with radius 2.5 ft, so the area of the top semicircle part is $A = \frac{1}{2} π {(2.5)}^{2} = 9.8 {ft}^{2} .$ Add the area of the rectangle to the area of the semicircle. Then, the total area to be covered with the window film is $A = 50 + 9.8 = 59.8 {ft}^{2} .$

Area within Area

Suppose you want to install a round hot tub on your backyard patio. How would you calculate the space needed for the hot tub? Or, let’s say that you want to purchase a new dining room table, but you are not sure if you have enough space for it. These are common issues people face every day. So, let’s take a look at how we solve these problems.

Example 14 Finding the Area within an Area

The patio in your backyard measures 20 ft by 10 ft (Figure 10.121). On one-half of the patio, you have a 4-foot diameter table with six chairs taking up an area of approximately 36 sq feet. On the other half of the patio, you want to install a hot tub measuring 6 ft in diameter. How much room will the table with six chairs and the hot tub take up? How much area is left over?

Figure 10.121

The hot tub has a radius of 3 ft. That area is then $A = π {(3)}^{2} = 9 π = 28.27 {ft}^{2} .$ The total square feet taken up with the table and chairs and the hot tub is $36 + 28.27 = 64.27 {ft}^{2} .$ The area left over is equal to the total area of the patio, $200 ft squared$ minus the area for the table and chairs and the hot tub. Thus, the area left over is $200 - 64.27 = 135.7 {ft}^{2} .$

People in Mathematics

Heron of Alexandria

Figure 10.122 Heron of AlexandriaHeron of Alexandria (credit: "Heron of Alexandria" from 1688 German translation of Hero's Pneumatics/Wikimedia Commons, Public Domain)

Heron of Alexandria, born around 20 A.D., was an inventor, a scientist, and an engineer. He is credited with the invention of the Aeolipile, one of the first steam engines centuries before the industrial revolution. Heron was the father of the vending machine. He talked about the idea of inserting a coin into a machine for it to perform a specific action in his book, Mechanics and Optics. His contribution to the field of mathematics was enormous. Metrica, a series of three books, was devoted to methods of finding the area and volume of three-dimensional objects like pyramids, cylinders, cones, and prisms. He also discovered and developed the procedures for finding square roots and cubic roots. However, he is probably best known for Heron’s formula, which is used for finding the area of a triangle based on the lengths of its sides. Given a triangle $A B C$ (Figure 10.123),

Figure 10.123

Heron’s formula is $A = \sqrt{s (s - a) (s - b) (s - c)},$ where $s$ is the semi-perimeter calculated as $s = \frac{a + b + c}{2} .$

Key Terms

triangle
square
rectangle
rhombus
apothem
radius
circle

Key Concepts

The area $A$ of a triangle is found with the formula $A = \frac{1}{2} b h,$ where $b$ is the base and $h$ is the height.
The area of a parallelogram is found using the formula $A = b h,$ where $b$ is the base and $h$ is the height.
The area of a rectangle is found using the formula $A = l w,$ where $l$ is the length and $w$ is the width.
The area of a trapezoid is found using the formula $A = \frac{1}{2} h (b_{1} + b_{2}),$ where $h$ is the height, $b sub 1$ is the length of one base, and $b sub 2$ is the length of the other base.
The area of a rhombus is found using the formula $A = \frac{d_{1} d_{2}}{2},$ where $d sub 1$ is the length of one diagonal and $d sub 2$ is the length of the other diagonal.
The area of a regular polygon is found using the formula $A = \frac{1}{2} a p,$ where $a$ is the apothem and $p$ is the perimeter.
The area of a circle is found using the formula $A = π r^{2},$ where $r$ is the radius.

Formulas

The area of a triangle is given as $A = \frac{1}{2} b h,$ where $b$ represents the base and $h$ represents the height.

The formula for the area of a square is $A = s \cdot s$ or $A = s^{2} .$

The area of a rectangle is given as $A = l w .$

The area of a parallelogram is $A = b h .$

The formula for the area of a trapezoid is given as $A = \frac{1}{2} h (a + b) .$

The area of a rhombus is found using one of these formulas:

$A = \frac{d_{1} d_{2}}{2},$ where $d sub 1$ and $d sub 2$ are the diagonals.
$A = \frac{1}{2} b h,$ where $b$ is the base and $h$ is the height.

The area of a regular polygon is found with the formula $A = \frac{1}{2} a p,$ where $a$ is the apothem and $p$ is the perimeter.

The area of a circle is given as $A = π r^{2},$ where $r$ is the radius.

Adapted from Contemporary Mathematics by OpenStax (openstax.org), licensed under CC BY-NC-SA 4.0. Changes were made. License: CC-BY-NC-SA-4.0.